Problem: Simplify the following expression: $y = \dfrac{-3x^2- 7x+40}{-3x + 8}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-3)}{(40)} &=& -120 \\ {a} + {b} &=& &=& {-7} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-120$ and add them together. Remember, since $-120$ is negative, one of the factors must be negative. The factors that add up to ${-7}$ will be your ${a}$ and ${b}$ When ${a}$ is ${8}$ and ${b}$ is ${-15}$ $ \begin{eqnarray} {ab} &=& ({8})({-15}) &=& -120 \\ {a} + {b} &=& {8} + {-15} &=& -7 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-3}x^2 +{8}x) + ({-15}x +{40}) $ Factor out the common factors: $ x(-3x + 8) + 5(-3x + 8)$ Now factor out $(-3x + 8)$ $ (-3x + 8)(x + 5)$ The original expression can therefore be written: $ \dfrac{(-3x + 8)(x + 5)}{-3x + 8}$ We are dividing by $-3x + 8$ , so $-3x + 8 \neq 0$ Therefore, $x \neq \frac{8}{3}$ This leaves us with $x + 5; x \neq \frac{8}{3}$.